Find an example of two nilpotent matrices over $\mathbb{R}^{4}$ whose minimal polynomials are equal, but that are not similar

Question

I've been trying to answer this question for a few hours now, but without luck. I'm struggling with coming up with suitable examples for nilpotent matrices in general. Any tips? And possibly an example satisfying the title's request?

Thanks in advance!

Answer 1

Hint: In a minimal polynomial, the exponent of a factor tells you the size of the largest Jordan for that eigenvalue. So, for example: if the minimal polynomial of $A$ is $(\lambda - 1)^2 (\lambda - 2)$, then the largest Jordan block for $1$ has size $2$.

A matrix is nilpotent iff its only eigenvalue is zero.

Two matrices are similar if and only if they have the same Jordan form.

Answer 2

$A=\pmatrix{0 & 0 &0&0\cr 1& 0&0&0\cr 0&0&0&0\cr 0&0&0&0}$

$B=\pmatrix{0 & 0 &0&0\cr 1& 0&0&0\cr 0&0&0&0\cr 0&0&1&0}$

$A^2=B^2=0$, their minimal polynomial is $P(X)=X^2$ the rank of $A$ is 1 and the rank of $B$ is 2, so they cannot be similar.